Wave Optics Test

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Wave Optics Test - 68

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Weekly Quiz Competition

Question 1
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The YDSE apparatus is as shown in the figure. The condition for point $$P$$ to be a dark fringe is ($$l=$$wavelength of light waves)

A
$$\left( { l }_{ 1 }-{ l }_{ 3 } \right) +\left( { l }_{ 2 }-{ l }_{ 4 } \right) =n\lambda $$

B
$$\left( { l }_{ 1 }-{ l }_{ 2 } \right) +\left( { l }_{ 3 }-{ l }_{ 4 } \right) =\cfrac { \left( 2n+1 \right) \lambda }{ 2 } $$

C
$$\left( { l }_{ 1 }+{ l }_{ 3 } \right) +\left( { l }_{ 2 }+{ l }_{ 4 } \right) =\cfrac { \left( 2n-1 \right) \lambda }{ 2 } $$

D
$$\left( { l }_{ 1 }-{ l }_{ 2 } \right) +\left( { l }_{ 4 }-{ l }_{ 3 } \right) =\cfrac { \left( 2n-1 \right) \lambda }{ 2 } $$

Solution

Path difference between $$SS_1P$$ and $$SS_2P$$ is
$$\Delta = (l_1 + l_3) -( l_2 + l_4) = ( n + \dfrac{1}{2})\lambda$$
$$\Rightarrow \Delta = (l_1 - l_2) + (l_3 -l_4) = ( n + \dfrac{1}{2})\lambda $$
Question 2
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A long horizontal slit is place $$1\ mm$$ above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen $$1\ m$$ away from the slit. If the mirror reflects only $$64\%$$ of the light falling on it, the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen is :

A
$$8 : 1$$

B
$$3 : 1$$

C
$$81 : 1$$

D
$$9 : 1$$

Solution

Before reflection, intensity of light is $$ {I}_{o}$$(say) and after reflection it becomes $$ {0.64I}_{o}$$
$$ { I }_{ max }={ I }_{ o }+{ 0.64I }_{ o }+2\sqrt { { I }_{ o }\times { 0.64I }_{ o } } =3.24{ I }_{ o }$$
$$ { I }_{ min }={ I }_{ o }+{ 0.64I }_{ o }-2\sqrt { { I }_{ o }\times { 0.64I }_{ o } } =0.04{I}_{o}$$
$$\frac { { I }_{ max } }{ { I }_{ min } } =\dfrac { 324 }{ 4 } =81:1$$
Question 3
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Directions For Questions

A narrow tube is bent in the form of a circle of radius $$R$$, as shown in figure. Two small holes $$S$$ and $$D$$ are made in the tube at the positions at right angle to each other. A source placed at $$S$$ generates a wave on intensity $${I}_{0}$$ which is equally divided into two parts; one part travels along the longer path, while the other travels along the shorter path. Both the waves meet at point $$D$$ where a detector is placed.

...view full instructions

If a maxima is formed at a detector, then the magnitude of wavelength $$\lambda$$ of the wave produced is given by

A
$$\pi R$$

B
$$\cfrac{\pi R}{2}$$

C
$$\cfrac{\pi R}{4}$$

D
all of these

Solution

Net path difference between the waves reaching at D, $$x= \dfrac{3}{2}\pi R- \dfrac{\pi R}{2}= \pi R$$
For maxima, $$\pi R= n \lambda$$
$$\lambda= \dfrac{\pi R}{n}$$
Thus $$\lambda= \pi R, \dfrac{\pi R}{2}, \dfrac{\pi R}{4} $$ and so on.
Question 4
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Directions For Questions

In a Young's double-slit experiment set up, source $$S$$ of wavelength $$6000\mathring { A } $$ illuminates two slits $${S}_{1}$$ and $${S}_{2}$$ which act two coherent sources. The source $$S$$ oscillates about its shown position according to the equation $$y=1+\cos { \pi t }$$, where $$y$$ is in millimeter and $$t$$ in second.

...view full instructions

At $$t=0$$, fringe width is $${\beta}_{1}$$, and at $$t=2s$$, fringe width of figure is $${\beta}_{2}$$. Then

A
$${\beta}_{1}> {\beta}_{2}$$

B
$${\beta}_{2}> {\beta}_{1}$$

C
$${\beta}_{1}= {\beta}_{2}$$

D
data is insufficient

Solution

The position of source at $$t=0s$$ is: $$y(0)=1+cos0=2mm$$
The position of source at $$t=2s$$ is: $$y(0)=1+cos2\pi=2mm$$
Hence, at the two different times, the position of the source is the same, and hence interference pattern formed is same and so is the fringe width.
Hence, $$\beta_1=\beta_2$$
Question 5
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Directions For Questions

A thin film of a specific material can be used to decrease the intensity of reflected light. There is destructive interference of waves reflected from upper and lower surfaces of the film. These films are called non-reflecting coatings. The process of coating the lens or surface with non-reflecting film is called blooming as shown in figure. The refracting index of coating ($${n}_{1}$$) is less than that of the glass ($${n}_{2}$$).

...view full instructions

If a light of wavelength $$\lambda$$ is incident normally and the thickness of film is $$t$$, then optical path difference between waves reflected from upper and lower surface of the film is

A
$${ 2n }_{ 1 }t$$

B
$${ 2n }_{ 1 }t-\cfrac { \lambda }{ 2 } $$

C
$${ 2n }_{ 1 }t+\cfrac { \lambda }{ 2 } $$

D
$$2t$$

Solution

Some part of the light gets reflected from the coating surface as light 1 and the other part gets refracted which further gets reflected from glass and after refraction ,emerges as light 2 and interfere with light 1.
Optical path the light 2 travelled from A to B and then From B' to A' be $$x$$.
Let $$t$$ be the thickness of the film.
Then $$x= 2 \times n_1t $$
Thus the total path difference is $$x= 2 n_1t $$
Question 6
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In a Young's double slit experiment set up, source $$S$$ of wavelength $$500 nm$$ illuminates two slits $$S_{1}$$ and $$S_{2}$$ which act as two coherent sources. The source $$S$$ oscillates about its own position according to the equation $$y = 0.5 \sin \pi t$$ where $$y$$ is in mm and $$t$$ in seconds. The minimum value of time $$t$$ for which the intensity at point $$P$$ on the screen exactly infront of the upper slit becomes minimum is :

A
$$1 s$$

B
$$2 s$$

C
$$3 s$$

D
$$1.5 s$$

Solution

By SHM, Max speed will be at mean position.
When source is moving towards Slit setup, Number of Photons Hitting the Point P will greater than the situation when the source is moving away from the setup.

Understanding by Droppler's Effect where the velocity of source changes the wavelength of sound.

Intensity will be lowest when the source is at mean position moving away from the setup.

For that
$$y = 0 = 0.5 \times \sin (\pi\times t)$$
t = Whole Numbers

Velocity of Source:
$$\dfrac{\mathrm{d} y}{\mathrm{d} t} = 0.5 \times \pi \times \cos(\pi\times t)$$

For Max Velocity:
$$\dfrac{\partial^2 { y}}{\partial t^2} = -0.5 \times \pi^{2} \times \sin(\pi\times t) = 0 $$
t = Whole Numbers
But for t = 1 , Velocity is max and in opposite direction.

Hence t = 1 is the answer
Question 7
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Directions For Questions

A narrow tube is bent in the form of a circle of radius $$R$$, as shown in figure. Two small holes $$S$$ and $$D$$ are made in the tube at the positions at right angle to each other. A source placed at $$S$$ generates a wave on intensity $${I}_{0}$$ which is equally divided into two parts; one part travels along the longer path, while the other travels along the shorter path. Both the waves meet at point $$D$$ where a detector is placed.

...view full instructions

If a minima is formed at the detector, then the magnitude of wavelength $$\lambda$$ of the wave produced is given by

A
$$2\pi R$$

B
$$\cfrac{3}{2}\pi R$$

C
$$\cfrac{5}{2}\pi R$$

D
none of these

Solution

Path difference at D, $$\Delta x= \pi R$$
For minima, $$\pi R= (n+\dfrac{1}{2}) \lambda$$
$$\implies \lambda= \dfrac{\pi R}{(n+\dfrac{1}{2})}$$
Thus, $$\lambda$$ can be $$ 2\pi R, \dfrac{2}{3}\pi R, \dfrac{2}{5}\pi R..........so on$$
Question 8
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Directions For Questions

Consider the situation shown in figure. The two slits $${S}_{1}$$ and $${S}_{2}$$ placed symmetrically around the central line are illuminated by monochromatic light of wavelength $$'\lambda\ '$$. The separation between the slits is $$'d'$$. The light transmitted by the slits falls on a screen $${S}_{0}$$ placed at a distance $$'D'$$ from the slits. The slit $${S}_{3}$$ is at the central line and the slit $${S}_{4}$$ is at a distance z from $${S}_{3}$$. Another screen $${S}_{c}$$ is placed a further distance $$'D\ '$$ away from $${S}_{c}$$. Find the ration of the maximum and minimum intensity observed on $${S}_{c}$$.

...view full instructions

If $$z=\cfrac{\lambda D}{d}$$

A
$$4$$

B
$$2$$

C
$$\infty$$

D
$$1$$

Solution

Intensity at:
$$S _{3} = 4 I _{o}$$
$$S _{4} = 4 I _{o}$$

$$I _{max} = 16I _{o}$$
$$I _{min} = 0$$

Hence $$\dfrac{I _{max}}{I _{min}} = \infty$$
Question 9
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Directions For Questions

In a Young's double-slit experiment set up, source $$S$$ of wavelength $$6000\mathring { A } $$ illuminates two slits $${S}_{1}$$ and $${S}_{2}$$ which act two coherent sources. The source $$S$$ oscillates about its shown position according to the equation $$y=1+\cos { \pi t }$$, where $$y$$ is in millimeter and $$t$$ in second.

...view full instructions

At $$t=2s$$, the position of central maxima is

A
$$2mm$$ above $$C$$

B
$$2mm$$ below $$C$$

C
$$4mm$$ above $$C$$

D
$$4mm$$ below $$C$$

Solution

Path Difference from Source:
$$\delta _{Source} = \dfrac{1\times 10^{-3}\times 2\times 10^{-3}}{1}$$

Path Difference on the screen:
$$\delta = \dfrac{x\times 2\times 10^{-3}}{2}$$

For Central Maxima:
$$\delta _{Source} = \delta$$

Which gives $$x = 2mm$$

And as the source is above, the central fringe will be below the central horizontal axis.

Answer is B: 2mm below C
Question 10
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Directions For Questions

The arrangement for a mirror experiment is shown in figure. $$'S'$$ is a point source of frequency $$6\times {10}^{14}Hz$$. $$D$$ and $$C$$ represent the two ends of a mirror placed horizontally and $$LOM$$ represents the screen.

...view full instructions

Determine the width of the region where the fringes will be visible

A
$$4cm$$

B
$$6cm$$

C
$$2cm$$

D
$$3cm$$

Solution

Rays emerging from point source S get reflected on the mirror DC at points D and C ,and then reach point L and N respectively on the screen.
LN is the region where the fringes will be visible.
In the similar triangle SDP and LDO, $$\dfrac{y_2}{1}=\dfrac{195}{5}$$
so $$y_2= 39 mm$$
In similar triangle SCP and NCO, $$\dfrac{y_1}{1} =\dfrac{190}{10}$$
so $$y_1= 19mm$$
$$y_2-y_1= 20 mm$$
Width of the region where the fringes will be visible is $$2 cm $$